Question

The second-order rate constant for the reaction A+ 2 B ->C + D is 0.34 dm' mo1·• s'. \\'hat is the concentration of C after (i) 20s, (ii) 15 mi n when the reactants are mixed with initial concentrations of [A] = 0.027 mol dm • 1 and [BJ= 0.130mol dm""?

Answer 1

Answer:

i) After 20 s the concentration of C is 0.024 M

ii) after 15 min the concentration of C is 0.027 M

Explanation:

Let´s assume that the reaction is first order for each reactive, making a second order global reaction. In that case:

v = k[A][B] = 0.34 M⁻¹ s⁻¹ * 0.027 M * 0.130 M = 1.2 x 10⁻³ M /s

First, let´s see how much time it takes for the reactants to disappear:

The rate of disappearance of A will be:

-Δ[A] / Δt = v

where:

Δ[A] = final - initial concentration of A

Δ[t] = elapsed time

Then:

Δ[t] = -Δ[A] / v

Δ[t] = - (-0.027 M) / 1.2 x 10⁻³ M /s = 22.5 s

The rate of disappearance of B will be:

-1/2Δ[B] / Δt = v

Δt = -1/2 * (-0.130 M) / 1.2 x 10⁻³ M /s

Δt = 54.2 s

Then, the reactant A will completely disappear in 22.5 s and it will limit the reaction.

The rate of production of C will be:

Δ[C] / Δt = v

where

Δ[C] = final concentration of C - initial concentration of C

Δt = final time - initial time

v = rate of the reaction

Then:

Δ[C] = v * Δt

Since the initial concentration of C is 0 and the initial time is 0, then:

[C] = v * t

i) t = 20 s

[C] =  1.2 x 10⁻³ M /s * 20 s = 0.024 M

ii) t = 15 min = 15 min * (60 s/ 1 min) = 900 s

The reaction does not occur beyond the 22.5 s which is the time in which A disappears. Then, the concentration of C at 15 min will be the same as the concentration at 22.5 s:

[C] = 1.2 x 10⁻³ M /s * 22.5 s = 0.027 M